Chem1000A Spring 2007 Practice Assignment 6 - Answers

Chem1000A Spring 2007
Practice Assignment 6 - Answers
1. Draw the Lewis structures of the following molecule (The first element is the central element).
Include formal charges in the Lewis structure and draw reasonable resonance structures. For the
determination of formal charges show the fragments (+ their charges) that result from formal
homolytic cleavage of all bonds.
(a) IO2F4 -
(b) IF6 +
(c) BCl3 (d) NO4 3-
(e) ClO4 -
(a) IO2F4 -
.. : O:
: ..
.. F F..
:
.. I ..
: .. F F..
:
: O..
:
-
(b) IF6 +
..
: F:
..
.. F..
:
..
:
+
.. F..
I F..
:
: F..
: F..
:
(c) BCl3
..
: Cl:
.. B ..
: Cl .. Cl:
..
(d) NO4 3-
..
: O :
.. N
O
O..
.. : O:
.. : -
-
+
-
:
- ..
(e) ClO4 -
: O..
: O:
Cl
O..
:
_:
O..
:
..
:
.. F
..
: .. F
: O:
.. Cl
: O
O:
_
..
..
: O:
.. -
: O:
..
F..
:
I ..
F..
:
: O:
: O..
.. _
: O:
Cl
O..
:
: O:
1-5
: O:
Cl ..
: O
O:
..
..
: O:
_

2. Looking at electronegativity differences, ∆χ, predict whether the following compounds are ionic or
covalent.
(a) TeF6 ∆χ = 3.98-2.10 = 1.88 covalent (polar)
(b) CF4 ∆χ = 3.98-2.55 = 1.43 covalent
(c) AlF3 ∆χ = 3.98-1.61 = 2.37 ionic
(d) NaF ∆χ = 3.98-0.93 = 3.05 ionic
(e) B2H6 ∆χ = 2.20-2.04 = 0.16 covalent
3. (a) Draw Lewis structures of the following molecules. (b) Specify electron-pair geometries and
molecular geometries of the following molecules according the VSEPR model. (c) Indicate the bond
dipole moments and the overall molecular dipole moments (if the molecule is polar). Specify whether
the molecule is polar or non-polar.
(i) ClO2 -
(ii) XeO4
(iii) BrF4 -
(iv) NO2F
(v) SeF4
(vi) SF3 +
(vii) OH2
(viii) OF2
(ix) CFCl3
(x) XeF2
(xi) SF5 -
(xii) XeO3F2
(xiii) PO2F2 -
(ii)
(i)
..
: Cl
: O..
:
-
: O :
Xe
O O
O..
.. : :
.. :
O..
:
..
..
.. +
: Cl ..
Cl
O :
: O
.. -
O O
..
electron-pair geometry: tetrahedral (SN = 4)
molecular geometry: bent(the two ClO bonds are equivalent)
polar molecule: molecular dipole moment is pointing towards the side of the oxygens, away from Cl
electron-pair geometry: tetrahedral
(SN =4)
molecular geometry: tetrahedral
bond dipole moments cancel each other,
therefore, no mol. dipole moment
XeO 4 is non-polar
(iii)
..
.. - .. F..
: ..
: F..
.. Br .. F
F
.. :
: ..
electron-pair geometry: octahedra (SN = 6)
molecular geometry: square planar
(lone pairs on opposite sides of the plane)
bond dipole moments cancel each other,
therefore, no molecular dipole moment
BrF 4 - is non-polar
2-5

(iv)
..
: .. F
.. : :
..
O:
N
O
-
S
F
..
F
F..
:
.. : :
: ..
.. +
..
O
..
F F : :
.. : :
.. ..
..
: .. F
.. -
: O:
N
O..
:
electron-pair geometry: trigonal planar
(both NO bonds are equivalent due to resonance)
molecular geometry: trigonal planar
The bond dipole moment of NF is significantly
larger than that of the NO bonds, hence,
the NF dipole is only partially compensated by the
NO dipoles. The resulting molecular dipole moment
points in the direction of the fluorine.
NOF 2 is polar.
(vi)
electron-pair geometry: tetrahedral
molecular geometry: trigonal pyramidal
The SF bond dipole moments point have all
a component pointing down.
SF 3 + is polar.
(viii)
electron-pair geometry: tetrahedral
molecular geometry: bent
The OF bond dipole moments point towards F.
F 2 O is polar.
+
+
F
N
3-5
O
O
+
(v)
.. ..
O
H H
Cl
Cl
.. : :
: ..
..
..
: F :
..
Cl .. :
..
: F :
F
Se
F
.. F:
:
..
.. :
: ..
.. :
+
electron-pair geometry: trigonal bipyramidal
(lone pair goes into the equatorial position)
molecular geometry: see saw
the axial SeF dipoles will cancel each other.
The equatorial SeF dipoles will add up to a
molecular dipole moment, pointing to the side
of the two equatorial fluorines.
SeF 4 is polar.
(vii)
electron-pair geometry: tetrahedral
molecular geometry: bent
The OH bond dipole moments point towards O.
H 2 O is polar.
(ix)
electron-pair geometry: tetrahedral
molecular geometry: tetrahedral
The CCl bond dipole moments are smaller than
that of the CF bond. Hence, the overall molecular
dipole moment points in the direction of F.
CFCl 3 is polar.
+
+

(x) (xi)
..
: F Xe .. F:
.. .. ..
.. ..
electron-pair geometry: trigonal bipyramidal
the three lone pairs are located in
the equatorial positions, therefore:
molecular geometry: linear
the XeF dipole moments cancel each other
XeF 2 is non-polar
(xii)
..
O..
..
: F :
Xe
..
F : :
..
O..
..
O..
electron-pair geometry: trigonal bipyramidal
molecular geometry: trigonal bipyramidal
the double-bond domains are located in the
equatorial plane.
The XeF bond dipole moments cancel each other.
The three XeO dipole moments cancel each other,
too. Therefore, XeO 3 F 2 is non-polar.
4-5
+
: ..
.. F
: F
..
..
..
: F : ..
F..
.. :
S F..
:
..
-
electron-pair geometry: octahedral
molecular geometry: square pyramidal
the four SF bond dipole moments of the
equatorial fluorines will cancel each other
the axial SF dipole remains
SF 5 - is a polar molecule.
(xiii)
:
..
.. :
..
O O
.. F
P
F
:
..
.. :
..
- ..
: :
..
:
..
: O
.. -
: O:
.. F
P
F
electron-pair geometry: tetrahedral
molecular geometry: tetrahedral
PF bond dipole moments are larger than those
of the PO bond, therefore, a small residual
dipole moment in the direction of the F's remains
PO 2 F 2 - is polar.
4. Determine the oxidation states in the following sulfur compounds (only one resonance structure is
given). Use the original definition of the oxidation state (by cleaving bonds). For (b) and (c),
distinguish between the oxidation states for individual sulfur atoms and an average oxidation state for
sulfur (i.e., the average of the two – or more – values). You would obtain these average oxidation
states if you use the ‘simple rules’, that are usually used for oxidation states.
(a) peroxodisulfate: S2O8 2-
+

-II
..
O..
-II
: O:
S
.. -I
O.. -I ..
O.. -II
: O:
-II ..
S O..
+VI : O..
:
-
-II
(b) S2F4
..
F
: S S F
: F F
: O..
:
-
-II
+VI
..
-I..
+I +III:
:
.. -I
.. :
.. : : .. :
-I -I
(c) disulfite S2O5 2-
-II -II
O:
O :
-II
: :
..
O..
S S:
+III
+V
O: .. : : O..
:
- -
-II -II
(d) dithionate S2O6 2-
-II -II
..
-IIO..
: O:
: O :
+V +V..
S S O..
-II
O: .. : : O..
:
- -
-II -II
5. Determine oxidation states for all elements in the following compounds:
(a) H-O-O-F H +I -O -I -O +I -F -I
(b) H3C-O-O-F H +I 3C -II -O -I -O +I -F -I
(c) glycol, H2C(OH)CH(OH)CH2(OH):
H
H
H
H
H
(d) ClO3 -
..
O.. ..
.. H
O
.. H
O..
H
H +I 2C -I (O -II H +I )C 0 H +I (O -II H +I )C -I H +I 2(O -II H +I )
Cl +V O -II 3 -
5-5